Given the function #f(x)= 1/12x^4 + 1/6x^3-3x^2-2x+1# how do you find any points of inflection and determine where the curve is concave up or down?

If #f# is twice differentiable on an open set A, the inflections are the points where #f^{''}# changes sing. This is a polynomial so it's obvious that it's smooth (i.e. infinitively differentiable) on #RR#, so here, the inflection points are the points where #f^{''}=0# and exist #n>1# such that #f^((2n+1))!=0#.

So #f^{''}=x^2 +x -6# and its roots are #{-1 +- sqrt(1+24)}/2 = {2, -3}#

Now we notice that the third derivative: #2x+1# has zero only in 1, so it's not zero in the candidate inflection points, and this tells us that -3 is a falling point and 2 is a rising point

(this does make sense since we know how the graphic of a quartic polynomial behave) Now, if the function is twice differentiable, the concavity is determined by the sign of its second derivative: #f^{''}>0 in U:= {x<-3 or x>2}# #f^{''}<0 in D :={-3 < x <2}#

To find points of inflection and determine where the curve is concave up or down for the function \( f(x) = \frac{1}{12}x^4 + \frac{1}{6}x^3 - 3x^2 - 2x + 1 \), follow these steps: 1. Find the second derivative of the function \( f(x) \), denoted as \( f''(x) \). 2. Set \( f''(x) = 0 \) and solve for \( x \) to find potential points of inflection. 3. Determine the sign of \( f''(x) \) around each potential point of inflection to identify where the curve is concave up or down. Now, let's go through these steps: 1. First derivative: \( f'(x) = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 6x - 2 \) 2. Second derivative: \( f''(x) = x^2 + x - 6 \) 3. Set \( f''(x) = 0 \) and solve for \( x \) to find potential points of inflection: \[ x^2 + x - 6 = 0 \] Solving this quadratic equation gives us \( x = -3 \) and \( x = 2 \). 4. Determine the sign of \( f''(x) \) around each potential point of inflection: - For \( x < -3 \), choose \( x = -4 \): \( f''(-4) = (-4)^2 + (-4) - 6 = 6 > 0 \) ⇒ Concave up. - For \( -3 < x < 2 \), choose \( x = 0 \): \( f''(0) = 0^2 + 0 - 6 = -6 < 0 \) ⇒ Concave down. - For \( x > 2 \), choose \( x = 3 \): \( f''(3) = 3^2 + 3 - 6 = 6 > 0 \) ⇒ Concave up. So, the function has points of inflection at \( x = -3 \) and \( x = 2 \), and it is concave up for \( x < -3 \) and \( x > 2 \), and concave down for \( -3 < x < 2 \).